By symmetry, you can get another one from the one you have. Think about Reference Angles.

By more symmetry, you can get two more from the one you missed. That makes five (5). Your graphing is exemplary.

Note: Check your Domain. You may not get all five (5).

Wooow, what a miss that was... Gotta watch for that in the future.

So with , we are at 4. Now by graphing I see that the other intersection point lies at . But I have no idea how to use symmetry to get that, as it doesn't seem to be an addition of any proper fraction of pi. Any hints on how to grab that last intersection algebraically?